Eng ISM Chapter 8 Statistics for engineering and sciences by mandenhall

7/22/2019 Eng ISM Chapter 8 Statistics for engineering and sciences by mandenhall

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Tests of Hypotheses 161

CHAPTER 8

..

Tests of Hypotheses

8.2 a. is the probability of rejecting a null hypothesis when it is true.

b. If the null hypothesis is rejected, the probability that the alternative hypothesis is not

correct is . By rejecting 0H , there is evidence to support aH , but we cannot

absolutely prove aH .

c. If + = 1, then = 1 = 1 P(Fail to reject 0H | 0H false) =P(reject 0H | 0H true).

But we know =P(Reject 0H | 0H true). However, it is true that as increases,

decreases and as decreases, increases.

8.4 a. = probability of rejecting 0H when 0H is true

= ( 8 if .2)P y p = = 1 7

0

( 7) 1 ( ) 1 .968 .032y

P y p y=

= = =

( ( 7)P y is found using Table 2, Appendix B, with n= 20 andp= .2)

b. = probability of rejecting 0H when 0H is true

= ( 5 if = .2)P y p = 1 4

0

( 4) 1 ( ) 1 .630 .370y

P y p y=

= = =


c. = probability of failing to reject 0H when it is false

= ( 7 if = .5)P y p =7

0

( ) .132y

p y=

=


d. = probability of failing to reject 0H when it is false

= ( 4 if = .5)P y p =4

0

( ) .006y

p y=

=


e. To minimize the probability of a Type I error, the rejection region 8y is more

desirable because it has a smaller value of . To minimize the probability of a Type II

error, the rejection region 5y is more desirable because it has a smaller value of .


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162 Chapter 8

f. = probability of rejecting 0H when 0H is true

= .01 =1 1

0 0

( if .2) 1 ( ) ( ) .990a a

y y

P y a p p y p y

= =

= = =

Using Table 2, Appendix B, with n= 20 andp= .20,

8

0

( ) .990 1 8 9y

p y a a=

= = =

The form of the rejection region is 9y for = .01

g. Power = 1 = 1 8

0

( 8 if .4) 1 ( ) 1 .596 .404y

P y p p y=

= = = =

( ( 8)P y is found using Table 2, Appendix, B, with n= 20 andp= .4)

h. Power = 1 = 1 8

0

( 8 if .7) 1 ( ) 1 .005 .995y

P y p p y=

= = = =

( ( 8)P y is found using Table 2, Appendix, B, with n= 20 andp= .7)

8.8 From Exercise 8.7,

2

1

( ) / 21( )

2

n

i

n y

L e

=

=

0( )

( )

L

L

=

Since 0 0 and ,y = =

2

21

2 21

1 12

2

1

( 0) / 2

/ 2/ 2 ( ) / 2

0

( ) / 2( ) / 2

1

( ) 2

( )

12

n

n

n ni

i

i in

i

n y

yy y y

y yn y y

eL e

e eL e

e

=

=

= =

=

= = = =

=

2 22 2 21 1 2 2 2

11 1 1

/ 2/ 2 / 2 / 2/ 2 / 2 ( ) / 2

nn n n

ii i i

y nyy y yny ny n ye e e e e e e== = =

= = =


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8.10 To determine if the egg hatching rate for a certain species of frog exceeds .5 when the eggs

are exposed to ultraviolet radiation, we test:

0: .5H p=

a: .5H p>

8.12 We define the following parameters:

p= true proportion of sevens occurring when two dice are rolled.

Since we are interested in testing whether this proportion differs from 1/6, we test:

0: 1/6H p=

a: 1/6H p

8.14 To determine if the mean amount of raidium-226 in soil in a Florida county exceeds 4pCi/L,we test:

0: 4H =

a: 4H >

8.16 a. Let = mean surface roughness of coated interior pipe used in oil fields. To determine

if this mean differs from 2 micrometers, we test:

0

a

: 2

: 2

H

H

=

b. Summary calculations yield the following:

Variable N Mean SD

ROUGHNESS 20 1.8810 0.5239

The test statistic is 01.8810 2

1.0160.5239 20

yz

s n

= = =

c. The rejection region for the small sample, two-tailed test requires /2 = .05/2 = .025 inboth tails of the tdistribution. From Table 7 in Appendix B, t.025= 2.093 with df = n

1 = 20 1 = 19. The rejection region isz> 2.093.

d. Since the observed value of the test statistic does not fall in the rejection region,H0cannot be rejected. There is insufficient evidence to indicate that the mean surface

roughness of coated interior pipe used in oil fields differs from 2 micrometers.

e. Both the confidence interval and test of hypothesis procedure give us information as to

where the value of the population mean is located. Both the confidence interval of

Exercise 7.24 and this test of hypothesis used a reliability level of 95%. They give

similar results.


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164 Chapter 8

8.18 To determine if the mean alkalinity level of water in the tributary exceeds 50 mpl, we test:

H0: = 50

Ha: > 50

The test statistic isz=0 67.8 50

14.4 / 100y

y

= = 12.36

The rejection region requires = .01 in the upper tail of thezdistribution. From Table 5,

Appendix B,z.01= 2.33. The rejection region isz> 2.33.

Since the observed value of the test statistic falls in the rejection region (z= 12.36 > 2.33),H0

is rejected. There is sufficient evidence to indicate that the mean alkalinity level of water in

the tributary exceeds 50 mpl at = .01.

8.20 We test: 0: 4H =

a: 4H <


3.889/ 2.081/ 26

yt

s n

= = =

The small sample one-tailed rejection region requires = .10 in the upper tail of the t

distribution with df = n= 1 = 26 1 = 25. From Table 7, Appendix B, .10t = 1.316. The

rejection region is t< 1.316.

Since the observed value of the test statistic does not fall in the rejection region (t= 3.889 1.440.

18,0002571.429

7

yy

n= = =

( )2

22

2

18,00046,365,200

79,485.7157 13,247.6191 6 6

yy

nsn

= = = =

13,247.619 115.098s= =

The test statistic is 02571.429 2500 71.429

1.6443.503/ 115.098 / 7

yt

s n

= = = =

Since the observed value of the test statistic falls in the rejection region (t= 1.64 >

1.440), 0H is rejected. There is sufficient evidence to indicate the mean breaking

strength is greater than 2500 pounds per linear foot at = .10. Yes, the pipe meets

specifications.

b. In terms of y :

0 0

115.0982,500 1.440 2,567.66

6

sy t

n

= + = + =

The rejection region in terms of y is y > 2,567.66

= probability of accepting 0H when it is false =P(y < 2,567.66 if a = 2,575)

To find this probability, findzwhen a= 2,575.

2,567.66 2,5750.156

115.098 6

ayzs n

= = =

Thus, ( 0.156)P t= < , with df = n1 = 6 1 = 5. From Table 7 in Appendix B,

> .10. The power of the test is 1 Power > .90


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166 Chapter 8

c. In terms of y :

0 0

115.0982,500 1.440 2,567.66

6

sy t

n

= + = + =

The rejection region in terms of y is y > 2,567.66

= probability of accepting 0H when it is false =P(y < 2,567.66 if a = 2,800)

To find this probability, findzwhen a= 2,800.

2,567.66 2,8004.945

115.098 6

ayzs n

= = =

Thus, ( 4.945)P t= < , with df = n1 = 6 1 = 5. From Table 7 in

Appendix B, .001 < < .005.

d. The power of the test is 1 .995 < Power < .999

8.26 Thep-values associated with each test statistic are found in Table 5, Appendix B.

a. p-value = ( 1.96) .5 (0 1.96) .5 .4750 .0250P z P z> = = =

b. p-value = ( 1.645) .5 (0 1.645) .5 .4500 .05P z P z> = = =

c. p-value = ( 2.67) .5 (0 2.67) .5 .4962 .0038P z P z> = = =

d. p-value = ( 1.25) .5 (0 1.25) .5 .3944 .1056P z P z> = = =

8.28 For the two-tailed test in Example 8.8,z= 4.03.

p-value =P(z> 4.03) +P(z> 4.03) = 2P(z> 4.03) 2(.0000) = 0

8.30 a. For the one-tailed test of Exercise 8.17,z= 5.471.

p-value =P(z> 5.471) 0

b. For the one-tailed test of Exercise 8.18,z= 12.36.

p-value =P(z> 12.36) 0

c. For the one-tailed test of Exercise 8.19,z= 2.33.

p-value =P(z< 2.33) = .5 - .4901 = .0099.

d. For the one-tailed test of Exercise 8.20, t = 3.889 with 25 degrees of freedom.

p-value =P(t< 3.889) < .0005


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e. For the one-tailed test of Exercise 8.21, t= 1.39 with 4 degrees of freedom.

p-value =P(t< 1.39) > .10

8.32 a. Let 1 = mean quality performance rating of competitive R&D contracts and 2 = mean

quality performance rating for sole source R&D contracts.

Since we wish to determine if the mean rating for the competitive contracts exceeds the

mean rating for the sole source contracts, we test:

0 1 2: 0H =

a 1 2: 0H >

b. The rejection region for the large-sample, one-tailed test requires = .05 in the upper

tail of thezdistribution. From Table 5, Appendix B, .05z = 1.645. The rejection region

isz> 1.645.

c. Since = .05 >p-value, we would reject 0H . There is sufficient evidence to indicate

the mean rating for competitive contracts exceeds the mean rating for sole source

contracts at = .05.

8.34 Thep-value for the test isp= .0054. At = .01, 0H is rejected. There is sufficient evidence

to indicate that the mean voltage readings of the two locations differ.

This is the same conclusion that we arrived at using the confidence interval procedure of

Exercise 7.40.

8.36 Let 1= mean relational intimacy score for participants in the CMC group and let 2= mean

relational intimacy score for participants in the FTF group. To compare the mean relationalintimacy score for participants in the two groups, we test:

0 1 2

a 1 2

: 0

: 0

H

H

=

The rejection region for the small sample, two-tailed test requires /2 = .10/2 = .05 in bothtails of the tdistribution with df = n1+ n2 2 = 24 + 24 2 = 46. From Table 7, Appendix B,

t.051.684. The rejection region is t> 1.684 or t< 1.684.

Assuming the population variances are equal, we find the pooled variance.

2 2 2 22 1 1 2 2

p

1 2

( 1) ( 1) (24 1)(.49 ) (24 1)(.38 )0.19225

2 (24 24 2)

n s n ss

n n

+ + = = =

+ +

The test statistic is 1 2 02 2

p p

1 2

( ) (3.54 3.53) 00.0790

0.19225 0.19225

24 24

y y Dt

s s

n n

= = =

++


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168 Chapter 8

Since the observed value of the test statistic does not fall in the rejection region (t= 0.0790 >/

1.684), 0H is rejected. There is insufficient evidence to indicate that the mean relational

intimacy score for participants in the two groups differ.

8.38 a. The statistical method appropriate for testing a difference in mean times between the

stacked and non-stacked menu displays is the independent samples comparison of

means test of hypothesis.

b. There are three necessary assumptions:

1) Both samples must be randomly and independently selected.

2) Both population of times must have approximate normal distributions.

3) The population variances for the stacked and non-stacked times must be equal.

c. Knowing only the sample means is not enough information to compare population

means. We also need sample standard deviations and sample sizes to make inferences

about population means.

d. Testing at any > .10 results in the fail to reject 0H conclusion. There is insufficient

evidence to indicate the mean time required differs for the stacked and non-stacked

menu displays.

8.40 a. To determine if the mean relational intimacy score for participants in the CMC group

will significantly increase between the first and third meetings, we test:

0 3 1

a 3 1

: 0

: 0

H

H

=

>

b. The data should be analyzed as matched pairs since all the participants in the study are

going to be measured twice, once after the first meeting and again after the thirdmeeting. The differences for the pairs of scores should be analyzed when comparing

the population means.

c. Thep-value for the test isp= .003. At = .01, 0H is rejected. There is sufficient

evidence to indicate that the mean relational intimacy score for participants in the CMC

group increased between the first and third meetings.

d. To determine if the mean relational intimacy score for participants in the FTF group

will significantly change between the first and third meetings, we test:

0 3 1

a 3 1

: 0

: 0

H

H

=

e. Thep-value for the test isp= .70. At = .01, 0H cannot be rejected. There is

insufficient evidence to indicate that the mean relational intimacy score for participants

in the FTF changed between the first and third meetings.


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8.42 a. To determine if the Huffman coding method will yield a smaller mean compression

ratio than the standard method, we test:

0

a

: 0

: 0

S H

S H

H

H

=

>

Summary information yields the following for the differences: 0.13d= and

d 0.1393s = .

The test statistic is 0

d

0.13 03.095

0.1393 11

d Dt

s n

= = = .

The rejection region requires = .05 in the upper tail of the tdistribution with df = n

1 = 11 1 = 10. From Table 7 in Appendix B, t.05= 1.812. The rejection region is t>

1.812.

Since the observed value of the test statistic does fall in the rejection region (z= 3.095> 1.812), 0H is rejected. There is sufficient evidence to indicate that the Huffman

coding method will yield a smaller mean compression ratio than the standard method.

b. These results do agree with the confidence interval conclusions reached in Exercise

7.46.

8.44 In order to compare means, we first must find the mean and standard deviation of the sample

differences.

Task Human Scheduler Automated Method Difference

1 185.4 180.4 5.0

2 146.3 248.5 102.23 174.4 185.5 11.1

4 184.9 216.4 31.5

5 240.0 269.3 29.36 253.8 249.6 4.2

7 238.8 282.0 43.2

8 263.5 315.9 52.4

260.532.56

8

dd

n

= = =

( )2

22

2d

( 260.5)17073.43

8 1227.271 8 1

dd

nsn

= = =

2d d 1227.27 35.03s s= = =

To determine if a difference exists between the mean throughput rates of human and

automated methods, we test:


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170 Chapter 8

0 d: 0H =

a d: 0H


d

32.56 02.629

/ 35.03 8

d Dt

s n

= = =

The rejection region requires /2 = .05/2 = .025 in both tails of the tdistribution with df =

n1 = 8 1 = 7. From Table 7, Appendix B, .05t = 2.365. The rejection region is t> 2.365

or t< 2.365.

Since the observed value of the test statistic falls in the rejection region (t= 2.629 1.96.

e. First, check to see if the normal approximation will be adequate:0 0

0 0 0

.07(.93)3 3 3 .07 3 .07 .060 (.010, .130)

163p

p qpqp p p

n n

Since the interval is completely in the interval (0, 1), the normal approximation will be

adequate.

Since the observed value of the test statistic does not fall in the rejection region

(z= 1.80 .05),H0is not rejected. This is the same

conclusion as in part e.

8.50 Letp= true mortality rate for adult rice weevils exposed to nitrogen. To determine if this

percentage exceeds 99%, we test:

0

a

: .99

: .99

H p

H p

=

>

The rejection region for the large sample, lower-tailed test requires = .10 in the upper tail ofthezdistribution. From Table 5 in Appendix B,z.01= 1.282. The rejection region isz>

1.282.

We first find31,386

.99888631,386 35

yp

n= = =

+


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172 Chapter 8


0 0

.998886 .9915.83

.99(.01) /31,421

p pz

p q n

= = =

Since the observed value of the test statistic does fall in the rejection region (z= 15.83 >

1.282), 0H is rejected. There is sufficient evidence to indicate that the true mortality rate for

adult rice weevils exposed to nitrogen exceeds 99%.

8.52 Letp= proportion of engineering Ph.D. degrees awarded to foreign nationals. Our sample

estimate is1,630

.72872,237

p= =

To determine if the true proportion of engineering Ph.D. degrees awarded to foreign nationals

exceeds .5, we test:

0: .5H p=

a: .5H p>


0 0

.7287 .521.63

.5(.5)

2,237

p pz

p q

n

= = =

Thep-value for the test is ( .21.63) 0p P z= > .

Using = .01 >p-value, we reject 0H . There is sufficient evidence to indicate the proportion

of engineering Ph.D. degrees awarded to foreign nationals exceeds .5.

8.54 For the Top of the Core:

First, check to see if the normal approximation is adequate:

p0 p3 p084

)5)(.5(.35.3 00

n

qp.5 .164 (.336, .664)

Since the interval falls completely in the interval (0, 1), the normal distribution will be

adequate.

64

84

yp

n= = = .762

To determine if the coat index exceeds .5, we test:

H0: p= .5

Ha: p> .5

The test statistic isz=

84

)5(.5.

5.762.

00

0 =

n

qp

pp= 4.80


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The rejection region requires = .05 in the upper tail of thezdistribution. From Table 5,

Appendix B,z.05= 1.645. The rejection region isz> 1.645.

Since the observed value of the test statistic falls in the rejection region (z= 4.80 > 1.645),H0

is rejected. There is sufficient evidence to indicate that the coat index exceeds .5 at = .05.

For the Middle of the Core:

First, check to see if the normal approximation is adequate:

p0 p3 p0n

qp 003 .5 73

)5)(.5(.3 .5 .176 (.324, .676)

Since the interval falls completely in the interval (0, 1), the normal distribution will be

adequate.

35

73

yp

n= = = .479

To determine if the coat index differs from .5, we test:

H0: p= .5

Ha: p.5


73

)5(.5.

5.479.

00

0 =

n

qp

pp= .36

The rejection region requires /2 = .05/2 = .025 in each tail of thezdistribution. From Table

5, Appendix B,z.025= 1.96. The rejection region isz< 1.96 orz> 1.96.

Since the observed value of the test statistic does not fall in the rejection region (z= .36


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174 Chapter 8


81

)5(.5.

5.358.

00

0 =

n

qp

pp= 2.56

The rejection region requires = .05 in the lower tail of thezdistribution. From Table 5,Appendix B,z.05= 1.645. The rejection region isz< 1.645.

Since the observed value of the test statistic falls in the rejection region (z= 2.56 < 1.645),H0is rejected. There is sufficient evidence to indicate that the coat index is less than .5 at

= .05.

8.56 a. Letp1= proportion of public wells with a detectable level of MTBE and letp2=

proportion of private wells with a detectable level of MTBE. To determine if theproportion of public and private well with a detectable level of MTBE differs, we test:

0 1 2

a 1 2

: 0

: 0

H p p

H p p

=

We first find 111

48 .40

120

yp

n= = = , 11

1

22 .2136

103

yp

n= = = , and

48 22 .3139

120 103p

+= =

+.

The test statistic is 1 2 0

1 2

( ) (.40 .2136) 02.99

1 11 1.3139(.6861)

120 103

p p Dz

pqn n

= = =

++

The rejection region for the large sample, two-tailed test requires /2 = .05/2 = .025 inboth tails of thezdistribution. From Table 5 in Appendix B,z.025= 1.96. The rejectionregion isz< 1.96 ofz> 1.96.


1.96), 0H is rejected. There is insufficient evidence to indicate that the proportion of

public and private well with a detectable level of MTBE differs.

b. Both techniques compare the two population proportions. Since they both wereanalyzed using the 95% reliability level, the results will be the same.

8.58 a. Letp1= proportion of caisson structures that are inactive and letp2= proportion of well

protector structures that are inactive. To determine if the proportion of caisson

structures that are inactive exceeds the proportion of well protector structures that areinactive, we test:

0 1 2

a 1 2

: 0

: 0

H p p

H p p

=

>


15/29


The rejection region for the large sample, two-tailed test requires = .10 in the uppertail of thezdistribution. From Table 5 in Appendix B,z.10= 1.282. The rejection

region isz>1.282.

We first find 1 2598 177 598 177

.5431, .4403, .51561,101 402 1,101 402

p p p +

= = = = = =+


1 2

( ) (.5431 .4403) 03.53

1 11 1.5156(.4844)

1,101 402

p p Dz

pqn n

= = =

++


1.282), 0H is rejected. There is sufficient evidence to indicate that the proportion of

caisson structures that are inactive exceeds the proportion of well protector structures

that are inactive.

b. Letp1= proportion of caisson structures that are inactive and letp2= proportion of

fixed platform structures that are inactive. To determine if the proportion of caisson

structures that are inactive exceeds the proportion of fixed platform structures that are

inactive, we test:

0 1 2

a 1 2

: 0

: 0

H p p

H p p

=

>

The rejection region for the large sample, two-tailed test requires = .10 in the uppertail of thezdistribution. From Table 5 in Appendix B,z.10= 1.282. The rejection

region isz> 1.282.

We first find 1 2598 450 598 450

.5431, .2372, .34961,101 1,897 1,101 1,897

p p p +

= = = = = =+


1 2

( ) (.5431 .2372) 016.93

1 11 1.3496(.6504)

1,101 1,897

p p Dz

pqn n

= = =

++

Since the observed value of the test statistic does fall in the rejection region (z= 16.93

> 1.282), 0H is rejected. There is sufficient evidence to indicate that the proportion of

caisson structures that are inactive exceeds the proportion of fixed platform structures

that are inactive.

c. Letp1= proportion of well protector structures that are inactive and letp2= proportion

of fixed platform structures that are inactive. To determine if the proportion of well

protector structures that are inactive differs from the proportion of fixed platform

structures that are inactive, we test:

0 1 2

a 1 2

: 0

: 0

H p p

H p p

=


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176 Chapter 8

The rejection region for the large sample, two-tailed test requires /2 = .10/2 = .05 inboth tails of thezdistribution. From Table 5 in Appendix B,z.05= 1.645. The rejection

region isz< 1.645 orz> 1.645.

We first find 1 2177 450 177 450

.4403, .2372, .2727402 1,897 402 1,897p p p

+

= = = = = =+


1 2

( ) (.4403 .2372) 08.30

1 11 1.2727(.7273)

402 1,897

p p Dz

pqn n

= = =

++


1.645), 0H is rejected. There is sufficient evidence to indicate that the proportion of

well protector structures that are inactive differs from the proportion of fixed platform

structures that are inactive.

8.60 Let 1p = the hatching rate of the sun-shaded eggs and 2p = the hatching rate of the unshaded

eggs. To determine a difference in these proportions, we test:

0 1 2: 0H p p =

a 1 2: 0H p p


1 2

( )

1 1

p p Dz

pqn n

=

+

where 1 234 31

.4857, .387570 80

p p= = = =

34 31 .4333, 1 .5667

70 80p q p

+= = = =

+

(.4857 .3875) 01.21

1 1.4333(.5667)

70 80

z

= =

+

The rejection region requires /2 = .05/2 = .025 in both tails of thezdistribution. From Table5, Appendix B, .025z = 1.96. The rejection region isz> 1.96 orz< 1.96.

Since the observed value of the test statistic does not fall in the rejection region (z= 1.21 >/

1.96), 0H is not rejected. There is insufficient evidence to indicate the hatching rates of the

sun-shaded and unshaded eggs differ.


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8.62 Let 1p = proportion of passive solar-heated homes that required less than 200 gallons of oil in

fuel consumption last year and 2p = proportion of solar-heated homes that required less than

200 gallons of oil in fuel consumption last year.

Since it is desired to determine if there is a difference between the proportions, we test:

0 1 2: 0H p p =

a 1 2: 0H p p

The sample proportions are:

1

37 .74,

50p = = 1 1 1 1 .74 .26q p= = =

2

46 .92

50p = = , 1 1 1 1 .92 .08q p= = =

( )( )

37 46 .83,

50 50p

+= =

+ 1 1 .83 .17q p= = =


1 2

( ) (.74 .92) 0 .182.40

.0751 11 1.83(.17)

50 50

p p Dz

pqn n

= = = =

++

The rejection region for a large-sample, two-tailed test with = .02 requires

/ 2 / 2 / 2( ) ( ) .02 (0 ) .5 .02 / 2 .5 .01 .4900P z z P z z P z z < + > = < < = = =

From Table 5, Appendix B, .01z = 2.33. The rejection region isz< 2.33 orz> 2.33.

Since the observed value of the test statistic falls in the rejection region (z= 2.40 < 2.33),

0H is rejected. There is sufficient evidence to indicate a difference between the proportions

1 2( 0)p p at = .02.

In order for the above test to be valid, 1 111

2

p qp

n and 2 22

2

2

qp

n must not contain 0

or 1.

1 11

1

.74(1 .74) 2 .74 2 .74 .124 (.616, .864)

50

p qp

n

2 22

2

.92(1 .92) 2 .92 2 .92 .077 (.8430, .997)

50

p qp

n

The above intervals do not contain 0 or 1, so the above test is valid.


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178 Chapter 8

8.64 Let 2 = variance of drill chip lengths.

2 20 0: 75 : 75H H = =

2 2a: 75 : 75aH H =

The test statistic is2 2

2

2 20

( 1) (50 1)(50.2) 21.95

(75)

n s

= =

The rejection region for the two-tailed test is 2 2 / 2 > or2 2

1 / 2 < where df = n1 = 50

1 = 49. From Table 8, Appendix B, with /2 = .05/2 = .025,2.025 71.4202 and

2.975 =

32.3574. The rejection region is2 2

71.4202 or 32.3574.> <

Since the observed value of the test statistic does fall in the rejection region2

0( 21.95 32.3574), H= < is rejected. There is sufficient evidence to indicate the standard

deviation of the drill chip lengths differ from 75 mm.

8.66 Let 2 = variance of the effluent turbidity in water specimens disinfected by the

preammoniation method.

20: .0016H =

2a: .0016H >

The test statistic is2 2

2

2

( 1) (44 1)(.16) 1.1008 688

.0016 .0016

n s

= = = =

The rejection region for a one-tailed test requires = .01 to be in the upper tail of the2

distribution with df = n1 = 44 1 = 43.

From Table 8, Appendix B, 2.01 63.6907. The rejection region is2

63.6907> .

Since the observed value of the test statistic falls in the rejection region2

0( 688 63.6907), H= > is rejected. There is sufficient evidence to indicate the variance

exceeds .0016 at = .01.

8.68 Let 2 = variance of the PCB readings.

( )2

22

2

41.7248.71

.29714297 .04951 7 1 6

yy

nsn

= = = =


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To determine if the variance of the PCB readings is less than .1, we test:

20: .1H =

2a: .1H <

The test statistic is2

2

2

( 1) (7 1).0495 .297 2.97

.1 .1

n s

= = = =

The rejection region for a one-tailed test requires = .05 in the upper tail of the2

distribution with df = n1 = 7 1 = 6. From Table 8, Appendix B,2.05 = 12.5916. The

rejection region is 2 12.5916.>

Since the observed value of the test statistic does not fall in the rejection region2( 2.97= 012.5916),H>/ is not rejected. There is insufficient evidence to indicate the

variance of the PCB readings is less than .1 at = .05.

8.70 a. Let 21 = the equality of heat rate variance for traditional gas turbines and22 = the

equality of heat rate variance for aeroderivative augmented gas turbines. To determine

if there is a difference in the variation of the two gas turbine types, we test:

21

0 22

21

a 22

: 1

: 1

H

H

=

The test statistic is

2

2Larger sample variance 2,651.9 4.297Smaller sample variance 1,279.3F= = =

The rejection region for two-tailed test requires /2 = .05/2 = .025 in the upper tail oftheFdistribution with numerator df = n1 1 = 7 1 = 6 and denominator df = n2 1 =

39 1 = 38. From Table 11 in Appendix B,F.0252.74. The rejection region isF>2.74.

Since the observed value of the test statistic does fall in the rejection region (F= 4.279

> 2.74), 0H is rejected. There is sufficient evidence to indicate that there is a

difference in the variation of the two gas turbine types.

This result casts doubt on the inferences made in Exercise 8.33. The requirement ofequal population variances is not supported by this test.


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180 Chapter 8

b. Let 21 = the equality of heat rate variance for advanced gas turbines and22 = the

equality of heat rate variance for aeroderivative augmented gas turbines. To determine

if there is a difference in the variation of the two gas turbine types, we test:

21

022

21

a 22

: 1

: 1

H

H

=


2

Larger sample variance 2,651.917.250

Smaller sample variance 638.51F= = =

The rejection region for two-tailed test requires /2 = .05/2 = .025 in the upper tail oftheFdistribution with numerator df = n1 1 = 7 1 = 6 and denominator df = n2 1 =

21 1 = 20. From Table 11 in Appendix B,F.025= 3.13. The rejection region isF>

3.13.

Since the observed value of the test statistic does fall in the rejection region (F= 17.25

> 3.13), 0H is rejected. There is sufficient evidence to indicate that there is a

difference in the variation of the two gas turbine types.

This result casts doubt on the inferences made in Exercise 8.33. The requirement of

equal population variances is not supported by this test.

8.72 Let 21 = variance of the one wet sampler readings and22 = variance of the three wet sampler

readings.

To determine if the variation in hydrogen readings for the two sampling schemes differ, wetest:

( )2 2 2 20 1 2 1 2: / 1H = =

( )2 2 2 2a 1 2 1 2: / 1H

The test statistic isF=2 212 22

Larger sample variance 6.35.871

Smaller sample variance 2.6

s

s= = =

The rejection region for a two-tailed test with = .05 requires:

/ 2 1 1( ) .025 with 1 365 1 364P F F n > = = = =

2 2and 1 365 1 364n = = =

From Table 11, Appendix B, .025F 1.00. The rejection region isF> 1.00.


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(because2 21 22 22 1

ands s

s scannot both be

greater than/ 2

F

at the same time)

Since the observed value of the test statistic falls in the rejection region (F= 5.871 > 1.00),

0H is rejected. There is sufficient evidence to indicate the variations in hydrogen readings

for the two sampling schemes differ at = .05.

8.74 Let 21 = population variance of the TOC levels at Bedford and22 = population variance of

the TOC levels at Foxcote.

Since it is desired to see if the TOC levels have a greater variation at Foxcote, we test:

( )2 2 2 20 1 2 1 2: / 1H = =

( )2 2 2 2a 1 2 1 2: / 1H < <

The test statistic isF=2 212 22

Larger sample variance 1.27 1.61291.75

Smaller sample variance .9216.96

s

s= = = =

The rejection region for a one-tailed test with = .05 requires:

1 1( ) .05 with 1 52 1 51P F F n > = = = =

2 2and 1 61 1 60n = = =


Since the observed value of the test statistic falls in the rejection region (F= 1.75 > 1.53), 0H

is rejected. There is sufficient evidence to indicate the TOC levels at Foxcote have greater

variation than those at Bedford at = .05.

8.76 From the hint,

/ 2

Larger sample variance

Smaller sample varianceP F

>

=2 21 2

/ 2 / 22 22 1

ors s

P F Fs s

> >

=2 21 2

/ 2 / 22 22 1

s sP F F

s s

> + >

=2 2

+ (from Exercise 8.75)

=


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182 Chapter 8

8.82

/

( )( )

!

( )

x nn ep x

x

eh

=

=

( ) ~ ( ) ( )g x p x h

/( )

!

x nn e e

x

( 1/ )

( 1/ )

~

~

x n

x n

e

e

+

+

( ) ~g x Gamma with * =x+ 1

* = /(n+ 1)

RejectH0ifP(> 0) >P(

0) using ( )g x dist

n

.

8.84 Using MINITAB, the descriptive statistics are:

Descriptive Statistics: Species

Variable Region N N* Mean SE Mean StDev Minimum Q1 Median

Species Dry Steppe 5 0 14.00 9.53 21.31 3.00 3.00 5.00

Gobi Desert 6 0 11.83 7.44 18.21 4.00 4.00 4.50

Variable Region Q3 Maximum

Species Dry Steppe 29.50 52.00

Gobi Desert 16.00 49.00

Let 1= average number of ant species found in the Dry Steppe and 2= average number ofant species found in the Gobi Desert.

2 2 2 22 1 1 2 2

1 2

( 1) ( 1) (5 1)21.31 (6 1)18.21386.0539

2 5 6 2p

n s n ss

n n

+ + = = =

+ +

To determine if there is a difference in the average number of ant species found at sites in the

two regions, we test:

H0: 1= 2

Ha: 12

The test statistic is 1 2

2

1 2

( ) (14 11.83) 00.18

1 11 1386.0539

5 6

o

p

y y Dt

sn n

= = =

++

The rejection region requires /2 = .05/2 = .025 in each tail of the t-distribution with df = n1+

n22 = 5 + 6 2 = 9. From Table 7, Appendix B, t.025= 2.262. The rejection region is t 2.262.


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Since the observed value of the test statistic does not fall in the rejection region (t= 0.18 >/

2.262),H0is not rejected. There is insufficient evidence to indicate a difference in the

average number of ant species found at sites in the two regions at = .05.

8.86 a. Letp1= proportion of female students who switched due to loss of interest in SME and

p2= proportion of male students who switched due to lack of interest in SME.

Some preliminary calculations are:

11

1

74

172

yp

n= = = .43; 22

2

72 .44;

163

yp

n= = = 1 23

1 2

74 72

172 163

y yp

n n

+ += =

+ += .436

To determine if the proportion of female students who switch due to lack of interest in

SME differs from the proportion of males who switch due to a lack of interest, we test:

H0: p1p2= 0Ha: p1p20


+

=

+

163

1

172

1)564(.436.

0)44.43(.

11

0)(

21

21

nnqp

pp= 0.18

The rejection region requires /2 = .10/2 = .05 in each tail of thezdistribution. From

Table 5, Appendix B,z.05= 1.645. The rejection region isz< 1.645 orz> 1.645.

Since the observed value of the test statistic does not fall in the rejection region

(z= 0.18


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184 Chapter 8

8.88 Let 21 = population variance of the readings of instrument 1 and22 = population variance of

the readings of instrument 2.

Since it is desired to determine if instrument 2 is less precise (more variable), we test:

( )2 2 2 2

0 1 2 1 2: / 1H = =

( )2 2 2 2a 1 2 1 2: / 1H < <

( )2

22 11

2 11

1

(145)4209

45 11 5 1 4

yy

ns

n

= = = =

( )2

22 22

2 22

2

(150)4540

405 101 5 1 4

yy

ns n

= = = =

The test statistic isF=2221

Larger sample variance 1010

Smaller sample variance 1

s

s= = =

The rejection region for a one-tailed test with = .05 requires:

1 2( ) .05 with 1 5 1 4P F F n > = = = =

2 1and 1 5 1 4n = = =


Since the observed value of the test statistic falls in the rejection region (F= 10 > 6.39), 0H

is rejected. There is sufficient evidence to indicate instrument 2 is less precise than

instrument 1 ( )2 21 2 < at = .05.

8.90 Let 1 = experimental mean ultimate torsion moment and 2 = theoretical mean ultimate

torsion moment.

Since it is desired to determine if the means differ, we test:

0 1 2: 0H =

a 1 2: 0H


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2.08.3467

6

dd

n= = =

( )2

22

2d

(2.08)2.2422

1.52116 .30421 6 1 5

dd

nsn

= = = =

d .3042 .5515s = =


d

.3467 0 .34671.54

.2251/ .5515 6

d Dt

s n

= = = =

The rejection region for a small-sample, two-tailed test requires /2 = .05/2 = .025 in each tail

of the tdistribution with df = n1 = 6 1 = 5.

From Table 7, Appendix B, .025t = 2.571 with 5 df. The rejection region is t< 2.571 and

t> 2.571.

Since the observed value of the test statistic does not fall in the rejection region (t= 1.54 / 2.571, 0H is not rejected. There is insufficient evidence to indicate

the experimental mean and theoretical mean differ at = .05.

8.92 Letp= proportion of healthy, non-pregnant women who become uncomfortably hot when

their core temperature reaches 40C. Since we want to see if the true proportion of healthy,

non-pregnant women who become uncomfortably hot when their core temperature reaches

40C is less than .75, we test:

0

: .75H p=

a: .75H p<

11 .46

24p= =

T-Beam

Difference

(Exper. Theor.)

1 .07

2 .55

3 .20

4 .205 .68

6 1.18


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186 Chapter 8


0 0

.46 .75 .293.28

.0884.75(.25)

24

p pz

p q

n

= = = =

The rejection region for a large sample, one-tailed test with = .10 requires

( ) .10 ( 0) .5 .10 .4000P z z P z z < = < < = =

From Table 5, Appendix B, .10z = 1.28. The rejection region isz< 1.28.

Since the observed value of the test statistic falls in the rejection region (z = 3.28 < 1.28),

0H is rejected. There is sufficient evidence to indicate the true proportion of healthy, non-

pregnant women who become uncomfortably hot when their core temperature reaches 40C

is less than .75 at = .10.

In order for the above test to be valid,

2 pq

pn

must not contain 0 or 1.

.46(.54).46 2 .46 .203 (.257, .663)

24

The interval does not contain 0 or 1, so the above test is valid.

8.94 Since it is desired to determine if the new process is more variable than the old and the

variance of the old process was .00156, we test:

20: .00156H =

2a: .00156H >


2

2

( 1) (100 1)(.00211) .20889 133.90

.00156 .00156

n s

= = = =

The rejection region for a one-tailed test requires = .05 in the upper tail of the2

distribution with df = n1 = 100 1 = 99. From Table 8, Appendix B,2.05 124.342.

The rejection region is 2 124.342> .

Since the observed value of the test statistic falls in the rejection region2

( 113.90 124.342),= > 0H is rejected. There is sufficient evidence to indicate the variancefor the new process is larger than the old at = .05.

8.96 a. Since it is desired to determine if the true mean inbreeding coefficient for this species

of wasp exceeds 0, we test:

0: 0H =

a: 0H >


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The rejection region for a large-sample, one-tailed test requires = .05 in the upper tail

of thezdistribution. From Table 5, Appendix B, .05z = 1.645. The rejection region isz

> 1.645.


.699/ .884 / 197

yz

s n

= = =

Since the observed value of the test statistic does not fall in the rejection region (z=

.699 >/ 1.645), 0H cannot be rejected. There is insufficient evidence to indicate the

mean inbreeding coefficient exceeds 0 at = .05.

b. The confidence interval, from Exercise 7.98 is (.06, .148). Since the interval contains

the value 0, we would make the same inference of fail to reject 0H .

8.98 a. Let 1 = mean experimental vapor pressure and 2 = mean calculated vapor pressure.

Since it is desired to determine if the mean difference differs from 0, we test:

0 1 2: 0H =

a 1 2: 0H

.011.0006875

16

dd

n= = =

( )2

22

2d

.011.003039

.0030316 .0002020961 16 1 15

dd

nsn

= = = =

Temperature

Difference

Exper. Calcul.

100.60 .006

101.36 .007

104.60 .015

106.44 .014

108.70 .022

110.96 .008112.62 .000

115.21 .002

116.69 .026

119.38 .029

121.08 .008

123.61 .000

124.90 .010

127.74 .010

130.24 .010

131.75 .010


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188 Chapter 8

d .000202096 .01422s = =


d

.0006875 0 .0006875.193

.003555/ .01422/ 16

d Dt

s n

= = = =

The rejection region for the small-sample, two-tailed test requires /2 = .05/2 = .025 ineach tail of the tdistribution with n1 = 16 1 = 15 df. From Table 7, Appendix B,

.025t = 2.131. The rejection region is t< 2.131 and t> 2.131.

Since the observed value of the test statistic does not fall in the rejection region (t=

.193 / 2.131, 0H is not rejected. There is insufficient evidence

to indicate the mean difference differs from 0 at = .05.

b. p-value = ( .193) ( .193) 2 ( .193)P t P t P t + =

2 ( .193) 2 ( 1.341)P t P t > p-value > 2(.10) p-value > .20

This result indicates that the probability of observing a tvalue at least as contradictory

to 0H as the one observed in part a(if 0H is in fact true) is greater than .20. We

would therefore fail to reject 0H for any preselected value of less than .20.

8.100 Let 1 = mean milk yield of cows that are given access to shade and 2 = mean milk yield of

cows that are not given access to shade.

Since it is desired to determine if there is a difference in the mean milk yields, we test:

0 1 2: 0H =

a 1 2: 0H

Assume 2 2 2 21 2 p 40 1600s s s= = = =


2p

1 2

( ) (367.4 330.8) 0 36.62.55

14.37591 11 11600

16 15

y y D

sn n

= = =

++

The rejection region for a small-sample, two-tailed test requires /2 = .10/2 = .05 in each tail

of the tdistribution with df = 1 2 2 16 15 2 29n n+ = + = . From Table 7, Appendix B, .05t =

1.699. The rejection region is t< 1.699 or t> 1.699.

Since the observed value of the test statistic falls in the rejection region (t= 2.55 > 1.699),

0H is rejected. There is sufficient evidence to indicate a difference between the mean milk

yields at = .10.


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8.102 Let 1 = mean day-long clear-sky solar radiation level in St. Joseph, MO, and 2 = mean day-

long clear-sky solar radiation level in Iowa Great Lakes.

To determine if the mean radiation levels differ, we test:

0 1 2

: 0H =

a 1 2: 0H

Using a matched pairs analysis, the printout gives us the test statistic of t= 11.765 and the

p-value of .0001.

For any > .0001, we can reject 0H . So, for = .001, we have sufficient evidence to

indicate the mean day-long clear-sky solar radiation level at the two sites differ.

Documents

Eng ISM Chapter 8 Statistics for engineering and sciences by mandenhall